# Strange limit: $\lim_{n\to \infty} \left(n-{\ln(n!)\over \sqrt[n]{n^k\ln2 \ln3\ln4\cdots\ln n}}\right)=2k$

Observing this limit $$\lim_{n\to \infty} \left(n-{\ln(n!)\over \sqrt[n]{n^k\ln2 \ln3\ln4\cdots\ln n}}\right)=2k$$

I did try a few quite large values of n; the limit seem to converge, but slow.

Can this limit be correct and how can I shows it?

• Note that $\ln(n!)$ is $(n-1)$ times arithmetic mean of $A=\{\ln 2,\ldots,\ln n\}$ and $\sqrt[n]{\ln2\ln3\cdots\ln n}$ is related to geometric mean of numbers in $A$ – Qurultay Mar 26 '18 at 18:20
• Not sure how you came up with this guess, but my numerical computation tells that the limit diverges to $-\infty$ regardless of the value of $k$. – Sangchul Lee Mar 26 '18 at 18:56
• Thank you @Sangchul Lee. – user545955 Mar 26 '18 at 19:12

## 1 Answer

Let me provide some computation that concludes the question. With a bit of effort, one can show that there exists a constant $C > 0$ satisfying

$$n - \frac{\log(n!)}{(n^k \prod_{j=2}^{n} \log j)^{1/n}} \leq - \operatorname{li}(n) - n \log\left(1 - \frac{1}{\log n}\right) + k \log n + \log\log n + C$$

for all $n \geq 2$ and $k \in \mathbb{R}$, where $\operatorname{li}(x)$ is the logarithmic integral function. Then it is not hard to check that this bound behaves like $-n^{1-o(1)}$ and hence the limit diverges.